show that the quderateral formed by joining the midpoint sides the rectangle is Rhombaz.
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let AB be =2x
therefore, AB=DC=2x
let AD be=2y
therefore, AD=BC=2y
therefore each four 90°triangle has,
hypotenuse=(x)^2+(y)^2
let (x)^2+(y)^2 be=a
therefore, all side has same sides making it a rhombus and if we join opposite mid-points the point where they join always produces 90° angle (in a rectangle); proving it a rhombus
therefore, AB=DC=2x
let AD be=2y
therefore, AD=BC=2y
therefore each four 90°triangle has,
hypotenuse=(x)^2+(y)^2
let (x)^2+(y)^2 be=a
therefore, all side has same sides making it a rhombus and if we join opposite mid-points the point where they join always produces 90° angle (in a rectangle); proving it a rhombus
aksa8:
sorry quadrilateral and Rhombus
ok
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